Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 121111;
const int M = 23;
vector<int> imerge(const vector<int> &a, const vector<int> &b) {
  vector<int> ret;
  int p_a = 0;
  int p_b = 0;
  int cnt = 0;
  while (cnt < 10) {
    if (p_a == 10)
      ret.push_back(b[p_b++]);
    else if (p_b == 10)
      ret.push_back(a[p_a++]);
    else if (a[p_a] < b[p_b])
      ret.push_back(a[p_a++]);
    else if (a[p_a] > b[p_b])
      ret.push_back(b[p_b++]);
    else {
      ret.push_back(a[p_a++]);
      p_b++;
    }
    cnt++;
  }
  return ret;
}
int height[N];
int anc[N][M];
vector<int> edges[N];
vector<int> people[N][M];
void dfs_get_anc(int cur, int last = -1) {
  if (last == -1) {
    height[cur] = 0;
    for (int i = 0; i < M; i++) anc[cur][i] = cur;
    for (int i = 1; i < M; i++) people[cur][i] = people[cur][0];
  } else {
    for (int i = 1; i < M; i++) {
      anc[cur][i] = anc[anc[cur][i - 1]][i - 1];
      people[cur][i] =
          imerge(people[cur][i - 1], people[anc[cur][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < edges[cur].size(); i++) {
    int v = edges[cur][i];
    if (v == last) continue;
    anc[v][0] = cur;
    height[v] = height[cur] + 1;
    dfs_get_anc(v, cur);
  }
}
void swim(int &x, int h) {
  for (int i = 0; h > 0; i++) {
    if (h & 1) x = anc[x][i];
    h /= 2;
  }
}
int get_lca(int x, int y) {
  if (height[x] > height[y]) swap(x, y);
  swim(y, height[y] - height[x]);
  if (x == y) return x;
  while (true) {
    int i;
    for (i = 0; anc[x][i] != anc[y][i]; i++)
      ;
    if (i == 0) return anc[x][0];
    x = anc[x][i - 1];
    y = anc[y][i - 1];
  }
  return -1;
}
vector<int> get_ans(int cur, int lca) {
  vector<int> ans = people[lca][0];
  for (int i = M - 1; i >= 0; i--)
    if (height[anc[cur][i]] >= height[lca]) {
      ans = imerge(ans, people[cur][i]);
      cur = anc[cur][i];
      if (cur == lca) break;
    }
  return ans;
}
int main() {
  ios ::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < 10; j++) people[i][0].push_back(999999);
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    people[x][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    vector<int> tmp = people[i][0];
    sort(tmp.begin(), tmp.end());
    people[i][0].clear();
    for (int j = 0; j < 10; j++) people[i][0].push_back(tmp[j]);
  }
  dfs_get_anc(1);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    int lca = get_lca(u, v);
    vector<int> ans = imerge(get_ans(u, lca), get_ans(v, lca));
    int cnt = 0;
    for (int i = 0; i < a; i++)
      if (ans[i] != 999999) cnt++;
    cout << cnt << ' ';
    for (int i = 0; i < a; i++)
      if (ans[i] != 999999) cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0;
}